A rock thrown straight up with a velocity of

29 m/s from the edge of a building just misses
the building as it comes down. The rock is
moving at 44 m/s when it strikes the ground.
How tall was the building? The acceleration of gravity is 9.8 m/s

Answer in units of m

To find the height of the building, we need to consider the motion of the rock both when it is thrown up and when it comes back down.

First, let's find the time it takes for the rock to reach its maximum height. We can use the formula:

v = u + at

where:
v = final velocity (0 m/s when the rock reaches its maximum height)
u = initial velocity (29 m/s)
a = acceleration (-9.8 m/s², considering the negative direction is up)
t = time

Rearranging the formula to solve for 't':

0 = 29 - 9.8t

Simplifying further:
9.8t = 29
t = 29 / 9.8
t ≈ 2.96 s

Therefore, it takes approximately 2.96 seconds for the rock to reach its maximum height.

Next, we can use the formula to find the maximum height (h) reached by the rock:

h = ut + (1/2)at²

where:
h = maximum height
u = initial velocity (29 m/s)
t = time (2.96 s)
a = acceleration (-9.8 m/s²)

Plugging in the values:
h = 29(2.96) + (1/2)(-9.8)(2.96)²
h = 85.84 - 43.856
h ≈ 41.984 meters

Therefore, the maximum height reached by the rock above the edge of the building is approximately 41.984 meters.

To find the height of the building, we need to consider the height from the maximum height to the ground. The rock will fall from this maximum height with a final velocity of 44 m/s.

Using the formula:

v² = u² + 2ah

where:
v = final velocity (44 m/s)
u = initial velocity (0 m/s when the rock reaches its maximum height)
a = acceleration (-9.8 m/s²)
h = height

Since the initial velocity (u) is 0 m/s, the equation simplifies to:

v² = 2ah

Plugging in the values:
44² = 2(-9.8)h
h = (44²) / (2(-9.8))
h ≈ 96.69 meters

Therefore, the height of the building is approximately 96.69 meters.